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If \(\begin{bmatrix} x & 3x-y\\[0.3em] 2x+z & 3y-\omega \\[0.3em] \end{bmatrix} \)\(=\begin{bmatrix} 3 & 2 \\[0.3em] 4 & 7 \\[0.3em] \end{bmatrix},\) find x,y,z,ω. |
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Answer» Given two matrices are equal, \(\begin{bmatrix} x & 3x-y\\[0.3em] 2x+z & 3y-\omega \\[0.3em] \end{bmatrix} \)\(=\begin{bmatrix} 3 & 2 \\[0.3em] 4 & 7 \\[0.3em] \end{bmatrix}\) We know that if two matrices are equal then the elements of each matrices are also equal. ∴x = 3 …(1) And 3x – y = 2 …(2) And 2x + z = 4 …(3) 3y – ω = 7 …(4) Putting the value of x in equation (2), 3 × 3 – y = 2 ⇒ 9 – y = 2 ⇒ y = 9 – 2 ⇒ y = 7 Now, putting the value of y in equation (4), 3×7 – ω = 7 ⇒ 21 – ω = 7 ⇒ ω = 21 – 7 ⇒ ω = 14 Again, Putting the value of x in equation (3), 2 × 3 + z = 4 ⇒ 6 + z = 4 ⇒ z = 4 – 6 ⇒ z = – 2 ∴ x = 3, y = 7, z = – 2 and ω = 14 |
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